# Calculate Numbers of Plate/Pipe/Strip Earthings (Part-2)

## (2)Calculate Number of Plate Earthing:

• The Earth Resistance of Single Plate electrode is calculated as per IS 3040:
• ### R=ρ/A√(3.14/A)

• Where ρ=Resistivity of Soil (Ω Meter),
• A=Area of both side of Plate (m2),
• Example: Calculate Number of CI Earthing Plate of 600×600 mm, System has Fault current 65KA for 1 Sec and Soil Resistivity is 100 Ω-Meters.
• Current Density At The Surface of Earth Electrode (As per IS 3043):
• Max. Allowable Current Density  I = 7.57×1000/(√ρxt) A/m2
• Max. Allowable Current Density  = 7.57×1000/(√100X1)=757 A/m2
• Surface area of both side of single 600×600 mm Plate= 2 x lxw=2 x 0.06×0.06 = 0.72 m2
• Max. current dissipated by one Earthing Plate = Current Density x Surface area of electrode
• Max. current dissipated by one Earthing Plate =757×0.72= 545.04 Amps
• Resistance of Earthing Plate (Isolated)(R)=ρ/A√(3.14/A)
• Resistance of Earthing Plate (Isolated)(R)=100/0.72x√(3.14/.072)=290.14 Ω
• Number of Earthing Plate required =Fault Current / Max.current dissipated by one Earthing Pipe.
• Number of Earthing Plate required= 65000/545.04 =119 No’s.
• Total Number of Earthing Plate required = 119 No’s.
• Overall resistance of 119 No of Earthing Plate=290.14/119=2.438 Ω.

## (3)Calculating Resistance of Bared Earthing Strip:

### 1)Calculation for earth resistance of buried Strip (As per IEEE):

• The Earth Resistance of  Single Strip of Rod buried in ground is
• ### R=ρ/Px3.14xL (loge (2xLxL/Wxh)+Q)

• Where ρ=Resistivity of Soil (Ω Meter),
• h=Depth of Electrode (Meter),
• w=Width of Strip or Diameter of Conductor (Meter)
• L=Length of Strip or Conductor (Meter)
• P and Q are Coefficients

### 2)Calculation for earth resistance of buried Strip(As per IS 3043):

• The Earth Resistance of  Single Strip of Rod buried in ground is
• ### R=100xρ/2×3.14xL (loge (2xLxL/Wxt))

• Where ρ=Resistivity of Soil (Ω Meter),
• L=Length of Strip or Conductor (cm)
• w=Width of Strip or Diameter of Conductor (cm)
• t= Depth of burial (cm)
• Example :
• Calculate Earthing Resistance of Earthing strip/wire of 36mm Diameter, 262 meter long buried at 500mm depth in ground, soil Resistivity is 65 Ω Meter.
• Here R = Resistance of earth rod in W.
• r = Resistivity of soil(Ω Meter) = 65 Ω Meter
• l = length of the rod (cm) = 262m = 26200 cm
• d = internal diameter of rod(cm) = 36mm = 3.6cm
• h = Depth of the buried strip/rod (cm)= 500mm = 50cm
• Resistance of Earthing Strip/Conductor (R)=ρ/2×3.14xL (loge (2xLxL/Wt))
• Resistance of Earthing Strip/Conductor (R)=65/2×3.14x26200xln(2x26200x26200/3.6×50)
• Resistance of Earthing Strip/Conductor (R)== 1.7 Ω

Jignesh Parmar has completed M.Tech (Power System Control), B.E(Electrical). He is member of Institution of Engineers (MIE) and CEng,India. Membership No:M-1473586.He has more than 16 years experience in Transmission -Distribution-Electrical Energy theft detection-Electrical Maintenance-Electrical Projects (Planning-Designing-Technical Review-coordination -Execution). He is Presently associate with one of the leading business group as a Deputy Manager at Ahmedabad,India. He has published numbers of Technical Articles in “Electrical Mirror”, “Electrical India”, “Lighting India”,”Smart Energy”, “Industrial Electrix”(Australian Power Publications) Magazines. He is Freelancer Programmer of Advance Excel and design useful Excel base Electrical Programs as per IS, NEC, IEC,IEEE codes. He is Technical Blogger and Familiar with English, Hindi, Gujarati, French languages. He wants to Share his experience & Knowledge and help technical enthusiasts to find suitable solutions and updating themselves on various Engineering Topics.

### 9 Responses to Calculate Numbers of Plate/Pipe/Strip Earthings (Part-2)

1. najeeb ansari says:

thank you sir…………

2. murali sir says:

thankq sir Calculation Numbers of Plate/Pipe/Strip Earthings ( it is very importent in the electrical system ) transmition distribution and utlity of electrical power also electrical switch geer
all high rais building ,sell phon tower,radio signal tower,radors tower and temple tower etc

3. Paul Lo says:

The earthing calculation seems incomplete without counting on the touch/ step potential rise due to the higest earth fault, specially in HV and MV distribution protection.

• Yes you are absolutely right..We should calculate Touch and Ground Potential for HV Earthing Distribution..
Please refer my earthing mat calculation sheet where I have consider for the Same and you have to look whether it is in within limit or Not.It is also applicable for accurate calculate for LV but for quick calculation for LV I think is it not so make big impact on Calculation..

4. Mohammed Kamel says:

Dear Parmar;
Really, all thanks for these esteemed efforts in most fields of electrical systems.
I need to leave my comment on this calculation, where I think the numbers of plates are basically depends on the heaviest short circuit current value at medium voltage level; in addition to the discriminated time for disconnection.
I think the noted short circuit value occurred at Main Distribution Board (MDB); this current will be discharged in the earthing network following the the lower resistance path until returning to the neutral point of the transformer connected to the MDB under concern; this loop will be broken by opening the incoming breaker to the faulty MDB in a specified time of 0.4 Sec. Hence, only the short circuit currents at MV level will dissipate in the earth mass via earth plates.

Regards,
Mohammed Kamel
Dar-Al Handasah (Shair&Parteners) Consultancy Group
Makkah- KSA

5. mobin says:

Sir,
how to measure earth resistance & resistivity of soil by using a megger ???

Also the earth resistance between electrodes??

Pls help me

6. jothilakshmi says:

dear sir,
how to calculate how many earthing electrode is need based on soil resistivity. any formula is there?kindly give the correct method of calculation..

7. JD.. says:

Dear sir,
What is value of P and Q are Coefficients in Calculating Resistance of Bared Earthing Strip.????

8. Bolarin says:

Could you please check your calculation in the example given. I think correct resistance as per As per IS 3043 method should be 0.626Ω and not 1.7 Ω.